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A transformer has 500 primary turns and 10 secondary turns. (a) If \(V_{p}\) is \(120 \mathrm{~V}(\mathrm{rms}),\) what is \(V_{s}\) with an open circuit? If the secondary now has a resistive load of \(15 \Omega,\) what is the current in the (b) primary and (c) secondary?

Short Answer

Expert verified
(a) 2.4 V, (b) 0.0032 A, (c) 0.16 A.

Step by step solution

01

Determine the Voltage Ratio

The voltage ratio in a transformer can be calculated using the formula:\[\frac{V_s}{V_p} = \frac{N_s}{N_p}\]Given that \(N_p = 500\) turns and \(N_s = 10\) turns, we have:\[\frac{V_s}{120} = \frac{10}{500}\] Simplifying this will give us the ratio for the voltages.
02

Calculate the Secondary Voltage

From the equation:\[\frac{V_s}{120} = \frac{1}{50}\]Solving for \(V_s\) gives:\[V_s = \frac{120}{50} = 2.4 \text{ V}\]Thus, the secondary voltage with an open circuit is \(2.4 \text{ V}\).
03

Compute the Secondary Current with Load

When a 15 \(\Omega\) resistive load is connected across the secondary, the current \(I_s\) can be found using Ohm's Law:\[I_s = \frac{V_s}{R} = \frac{2.4}{15}\]Solving for \(I_s\) gives:\[I_s = 0.16 \text{ A}\]
04

Calculate the Primary Current

The power delivered to the secondary equals the power drawn from the primary:\[P_p = P_s\]Thus,\[V_p \times I_p = V_s \times I_s\]Substituting the known values:\[120 \times I_p = 2.4 \times 0.16\]Solving for \(I_p\), we find:\[I_p = \frac{2.4 \times 0.16}{120} = 0.0032 \text{ A}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Voltage Ratio
In transformer operations, understanding the voltage ratio is crucial. It connects the number of turns in the primary and secondary coils to the voltages across these coils. Transformers work on the principle of electromagnetic induction, where the voltage ratio is expressed as:
  • \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \)
This equation tells us that the ratio of the secondary voltage \(V_s\) to the primary voltage \(V_p\) is the same as the ratio of the number of turns in these coils, \(N_s\) and \(N_p\). In practical applications, this ratio governs how electrical energy is transformed between different voltage levels. This exercise involves a transformer with 500 primary turns and 10 secondary turns.By applying the voltage ratio, you can see that both the voltages and the turns share a common ratio of 1:50, meaning the secondary voltage is considerably lower.
Ohm's Law
Ohm's Law is fundamental when dealing with electrical circuits. It relates voltage, current, and resistance in a linear relationship:
  • \( V = IR \)
where\( V \) is the voltage across the component, \( I \) is the current flowing through it, and \( R \) is the resistance of the component.In the case of the secondary circuit in a transformer, once the secondary voltage is known, Ohm's Law can help calculate the current flowing through a given resistive load.
  • \( I_s = \frac{V_s}{R} \)
Here, we determined that the secondary voltage \(V_s\) was 2.4V. Plugging this into Ohm's Law with a resistive load of 15 \(\Omega\) allows calculation of the secondary current \(I_s\), resulting in a flow of 0.16 A through the load.
Primary and Secondary Circuits
Transformers operate by transferring electrical energy between two distinct circuits: the primary and secondary. These circuits include coils of wire wrapped around a magnetic core. The primary circuit is connected to a power source, whereas the secondary circuit is linked to a load.
  • Primary Circuit: Where the transformer's input voltage \(V_p\) is applied. It consists of turns \(N_p\) that create a magnetic field when current flows through.
  • Secondary Circuit: This circuit obtains energy through a magnetic field from the primary. Its voltage \(V_s\), and current \(I_s\) depend upon loads connected to it, like resistors.
Energy transfer occurs without direct electrical connections, purely through a magnetic field. The power in these circuits fulfills the conservation of energy, where the input power equals the output power, barring losses.
Resistive Load
A resistive load is any component in the circuit that consumes power by converting electrical energy into heat. Everyday examples include light bulbs and electric heaters. When a resistive load is connected to the secondary circuit of a transformer, it influences the current flow where Ohm's Law is heavily relied upon.Given a transformer setup, knowing the resistive load, like in this exercise with a 15 \(\Omega\) load, makes it possible to compute the exact current flowing into the load. The formula \( I = \frac{V}{R} \) captures this relationship, showing that current varies inversely with resistance for a given voltage.The resistive load is crucial because it determines the amount of energy absorbed and utilized within the secondary circuit. It plays a pivotal role in informing how a transformer is utilized in practical scenarios, affecting both efficiency and performance.

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Most popular questions from this chapter

In an oscillating \(L C\) circuit, \(L=8.00 \mathrm{mH}\) and \(C=1.40 \mu \mathrm{F}\). At time \(t=0,\) the current is maximum at \(12.0 \mathrm{~mA}\). (a) What is the maximum charge on the capacitor during the oscillations? (b) At what earliest time \(t>0\) is the rate of change of energy in the capacitor maximum? (c) What is that maximum rate of change?

In an oscillating \(L C\) circuit, \(L=3.00 \mathrm{mH}\) and \(C=2.70 \mu \mathrm{F}\). At \(t=0\) the charge on the capacitor is zero and the current is \(2.00 \mathrm{~A}\). (a) What is the maximum charge that will appear on the capacitor? (b) At what earliest time \(t>0\) is the rate at which energy is stored in the capacitor greatest, and (c) what is that greatest rate?

In an oscillating \(L C\) circuit, when \(75.0 \%\) of the total energy is stored in the inductor's magnetic field, (a) what multiple of the maximum charge is on the capacitor and (b) what multiple of the maximum current is in the inductor?

In an oscillating \(L C\) circuit in which \(C=4.00 \mu \mathrm{F},\) the maximum potential difference across the capacitor during the oscillations is \(1.50 \mathrm{~V}\) and the maximum current through the inductor is \(50.0 \mathrm{~mA}\). What are (a) the inductance \(L\) and (b) the frequency of the oscillations? (c) How much time is required for the charge on the capacitor to rise from zero to its maximum value?

An ac generator produces emf \(\mathscr{E}=\mathscr{E}_{m} \sin \left(\omega_{d} t-\pi / 4\right),\) where \(\mathscr{E}_{m}=30.0 \mathrm{~V}\) and \(\omega_{d}=350 \mathrm{rad} / \mathrm{s} .\) The current in the circuit attached to the generator is \(i(t)=I \sin \left(\omega_{d} t+\pi / 4\right),\) where \(I=620 \mathrm{~mA}\). (a) At what time after \(t=0\) does the generator emf first reach a maximum? (b) At what time after \(t=0\) does the current first reach a maximum? (c) The circuit contains a single element other than the generator. Is it a capacitor, an inductor, or a resistor? Justify your answer. (d) What is the value of the capacitance, inductance, or resistance, as the case may be?

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